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A constant force applied to a 2.4 -kg book produces acceleration \(3.4 \mathrm{~m} / \mathrm{s}^{2} \hat{\imath}-2.8 \mathrm{~m} / \mathrm{s}^{2} \hat{\jmath}\). What acceleration would result with a \(3.6-\mathrm{kg}\) book subject to the same force?

Short Answer

Expert verified
The acceleration for the 3.6 kg book is \( 2.27 \hat{\imath} - 1.87 \hat{\jmath} \) m/s².

Step by step solution

01

Determine the Force

First, use Newton's second law, which states that \( F = ma \). The force can be calculated by reorganizing the equation. Here, \( m = 2.4 \) kg and \( \vec{a} = 3.4 \mathrm{~m/s^2} \hat{\imath} - 2.8 \mathrm{~m/s^2} \hat{\jmath} \). Therefore, \( \vec{F} = 2.4 \times (3.4 \hat{\imath} - 2.8 \hat{\jmath}) \).
02

Calculate the Force Components

Compute the force vector components: \( F_x = 2.4 \times 3.4 = 8.16 \) N and \( F_y = 2.4 \times (-2.8) = -6.72 \) N. So, \( \vec{F} = 8.16 \hat{\imath} - 6.72 \hat{\jmath} \) N.
03

Apply the Force to the New Mass

For the new mass \( m = 3.6 \) kg, calculate the acceleration using \( \vec{a} = \frac{\vec{F}}{m} \): \( a_x = \frac{8.16}{3.6} \) and \( a_y = \frac{-6.72}{3.6} \).
04

Calculate Acceleration Components

Calculate each component: \( a_x = 2.27 \) m/s² and \( a_y = -1.87 \) m/s². Thus, the acceleration vector is \( \vec{a} = 2.27 \hat{\imath} - 1.87 \hat{\jmath} \) m/s².

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Force
When we talk about force, we're referring to something that can push or pull an object, causing it to start moving, stop, or change its speed or direction. In this context, it's important to understand that force is a vector, meaning it has both magnitude and direction. Force is often described using Newton's second law of motion, which links force (F) to mass (m) and acceleration (a) using the equation \( F = ma \). This equation tells us that force results in acceleration, and this change in motion depends on how massive the object is. The greater the force applied, the more acceleration it will cause, and conversely, for the same force, a heavier object will accelerate less than a lighter one.
In our exercise, we calculated the force applied to a book by multiplying its mass (2.4 kg) by its acceleration, providing us with the force in vector form. This allows us to explore how this same force would affect another object with different mass.
Acceleration
Acceleration is the measure of how quickly an object speeds up, slows down, or changes direction. Like force, acceleration is also a vector quantity, meaning it not only has a magnitude (how much) but also a direction (where to). This gives acceleration a form that can be illustrated as vector components, usually expressed in terms of units like meters per second squared (m/s²). In the problem, the book's acceleration was expressed in two-dimensional components: \(3.4 \, \mathrm{m/s^2} \, \hat{\imath}\) for the horizontal direction, and \(-2.8 \, \mathrm{m/s^2} \, \hat{\jmath}\) for the vertical direction.
Understanding acceleration as a vector helps in grasping how different components can be independently calculated and then combined to get the overall effect on object motion. Once the force acting on an object is known, the acceleration can be derived for any object regardless of its mass by rearranging Newton's second law to \( a = \frac{F}{m} \). This is precisely how we calculated the new acceleration for a book of different mass in the exercise.
Vector Components
In physics, vectors are essential for representing quantities that have both magnitude and direction, such as force and acceleration. Vector components break down a vector into parts that are easier to work with, typically along perpendicular axes like the x-axis (horizontal) and y-axis (vertical). This simplification is crucial when dealing with two-dimensional motion as it allows individual vector components to be computed and analyzed separately.
In our step-by-step problem, we separated the force and acceleration into their x (i) and y (j) components to make calculations straightforward. By doing so, we can understand how the object moves in each direction separately and then combine these movements to understand the total motion. This approach is particularly useful in solving real-world problems where forces and accelerations are rarely aligned with the neat, straightforward path of an axis.
Mass
Mass is a fundamental property of physical objects, indicating the amount of matter present. It is usually measured in kilograms (kg) and is invariant; meaning it doesn’t change regardless of the object's location in the universe. Mass plays a critical role in Newton's second law; the greater the mass of an object, the greater the force needed to achieve the same acceleration. This can be understood through the relationship \( F = ma \), where for a constant force, an increase in mass results in a decrease in acceleration.
In the exercise, we started with a known mass and acceleration to deduce the force exerted. This calculated force was then used for another object with a different mass, showing us how the acceleration changes with mass. This illustrates that while mass doesn’t dictate the direction of motion, it does heavily influence how that motion changes under applied forces.

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Most popular questions from this chapter

A 670 -kg helicopter rises straight up with acceleration \(1.20 \mathrm{~m} / \mathrm{s}^{2}\). (a) What upward force must the helicopter's rotor provide? (b) The helicopter then begins its descent with downward acceleration \(1.20 \mathrm{~m} / \mathrm{s}^{2} .\) Now what force does the rotor provide? Explain why your answers differ.

Experiments show that a walking person's hips describe circular arcs centered on the point of contact with the ground, and having radii equal to the leg's length \(L\) (see Figure \(\mathrm{P} 4.105\) ). Since the person's center of mass (more in Chapter 6 ) is near the hip, we can model the walker as a mass \(M\) moving in a circular arc of radius \(L\). In this case, \(M\) is the mass above the hip, which is roughly the person's total mass. At maximum speed, gravity alone is sufficient to provide the centripetal force. (a) Apply Newton's second law and show that the maximum speed at which a person can walk, according to this model, is \(v_{\max }=\sqrt{L g}\). (To move faster, one must run.) (b) What's the fastest walking speed for a typical \(75-\mathrm{kg}\) adult male? Use measurements on yourself or a friend to determine \(L\).

An 810 -kg compact car rounds a level curve at \(25 \mathrm{~m} / \mathrm{s}\). A \(2430-\mathrm{kg}\) SUV rounds the same curve at \(12.5 \mathrm{~m} / \mathrm{s}\). Compared with the car, the SUV's centripetal acceleration is (a) the same; (b) three times; (c) one-third; (d) one-half; (e) one-fourth.

A piece of steel rests on a wooden board, with frictional coefficient 0.35 between steel and wood. Find the maximum inclination angle for the board before the steel begins sliding.

A 0.17 -kg cue ball rests on the pool table. It's struck by a cue stick applying force \(\vec{F}_{1}=15 \mathrm{~N} \hat{\imath}+36 \mathrm{~N} \hat{\jmath}\). (a) Determine the magnitude and direction of the ball's acceleration. (b) If the force was applied for \(0.015 \mathrm{~s}\), what's the ball's final speed?

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